Given $n\in\Bbb N$, $n > 2$, find the reduction formula for:
$$
\int \cosh^n x dx
$$
I've derived an answer, but the signs do not match, so my question would be: where is the mistake in my calculation?
I'll use integration by parts:
$$
u = \cosh^{n-1}x \\
du = (n-1)\cosh^{n-2}x\sinh x\ dx\\
dv = \cosh x\ dx\\
v = \sinh x
$$
Thus the integral becomes:
$$
\begin{align}
J_n &= uv – \int vdu \\
&= \cosh^{n-1}x\sinh x – (n-1)\int \sinh^2x\cosh^{n-2}x\ dx\\
&= \cosh^{n-1}x\sinh x – (n-1)\left(\int(\cosh^2x-1)\cosh^{n-2}x\ dx\right)\\
&= \cosh^{n-1}x\sinh x – (n-1)(J_n – J_{n-2}) \iff \\
J_n &= \frac{\cosh^{n-1}x\sinh x + (n-1)J_{n-2}}{n}
\end{align}
$$
However, the answer section suggests that (note the minus sign):
$$
J_n = \color{red}{-}\frac{\cosh^{n-1}x\sinh x}{n} + \frac{n-1}{n}J_{n-2}
$$
I don't see why the first term is negative. Did I make a mistake somewhere or is it a typo in the book?
Thank you!
Best Answer
Your solution is completely correct! Definitely a typo in your book. Checked my own table of integrals and the one from Wikipedia; both agree with you. Well done!